An integrable and periodic function $f(x)$ satisfies $\int_{0}^{T}f(x)dx=\int_{a}^{a+T}f(x)dx$.

I want to prove:

For an integrable function $f(x)$ and periodic with period $T$, for every $a \in \mathbb{R}$, $$\int_{0}^{T}f(x)\;dx=\int_{a}^{a+T}f(x)\;dx.$$

I tried to change the values and define $y=a+x$ so that $dy=dx$ and the limits of the integrals are as we want, but I'm not sure how to use the fact that $f(x)$ is periodic.

Thanks a lot!

Solution 1:

$$ \begin{align} \int_a^{a+T}f(x)\,\mathrm{d}x-\int_0^{T}f(x)\,\mathrm{d}x &=\left(\color{red}{\int_a^{T}f(x)\,\mathrm{d}x}+\int_T^{a+T}f(x)\,\mathrm{d}x\right)\\ &-\left(\int_0^{a}f(x)\,\mathrm{d}x+\color{red}{\int_a^{T}f(x)\,\mathrm{d}x}\right)\\ &=\int_T^{a+T}f(x)\,\mathrm{d}x-\int_0^{a}f(x)\,\mathrm{d}x\\ &=\int_0^{a}f(x+T)\,\mathrm{d}x-\int_0^{a}f(x)\,\mathrm{d}x\\ &=\int_0^{a}(f(x+T)-f(x))\,\mathrm{d}x\\ &=\int_0^{a}0\,\mathrm{d}x\\ &=0 \end{align} $$

Solution 2:

If $F$ is a primitive of $f$, then

$$\int_{a}^{a+T}f(x)\ dx-\int_{0}^{T}f(x)\ dx$$ $$=F(a+T)-F(a)-F(T)+F(0)$$ $$=\Big(F(a+T)-F(T)\Big)-\Big(F(a)-F(0)\Big)$$ $$=\int_T^{a+T}f(x)\ dx-\int_0^af(x)\ dx$$ $$=0.$$

One checks the last equality by making the obvious change of variable, and by using the periodicity.

EDIT 1. What I wrote above is how I remember the computation. Of course, it can be written like that: $$ \int_{a}^{a+T}f(x)\ dx-\int_{0}^{T}f(x)\ dx=\int_T^{a+T}f(x)\ dx-\int_0^af(x)\ dx=0. $$

EDIT 2. Formal justification of the first equality in the above display: $$ \int_0^af(x)\ dx+\int_{a}^{a+T}f(x)\ dx=\int_{0}^{T}f(x)\ dx+\int_T^{a+T}f(x)\ dx. $$

(This formula should appear somewhere...)